The equation of plane passing through the point of (3, 2, 0) and containing the line \(\displaystyle\frac{x-2}{2}=\frac{y+3}{4}=\frac{z-1}{1}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(3x - y - 2z = 7\)
The line passes through A(2, −3, 1) with direction \(\vec{b} = (2, 4, 1)\). The plane also contains B(3, 2, 0), so it contains
\(\overrightarrow{AB} = (1, 5, -1)\)
The normal to the plane is perpendicular to both, so take their cross product:
\( \displaystyle \vec{n} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\\ 2 & 4 & 1\\ 1 & 5 & -1 \end{vmatrix} \)
\(= (-4 - 5)\hat{\imath} - (-2 - 1)\hat{\jmath} + (10 - 4)\hat{k}\)
\(= -9\hat{\imath} + 3\hat{\jmath} + 6\hat{k}\)
Dividing by −3, the normal is (3, −1, −2). The plane through B(3, 2, 0):
\(3(x - 3) - 1(y - 2) - 2(z - 0) = 0\)
\(3x - y - 2z = 7\)
Check with A(2, −3, 1): 6 + 3 − 2 = 7.
Note: This topic was not in the CUET 2026 syllabus.